Survival's Differential Equation under a Force

Why Survival Has a Differential Equation

A survival function describes the probability that a life remains alive beyond a specified duration. The force of mortality describes the instantaneous rate of death among lives who have survived to that duration. These quantities are connected: the force determines how quickly survival is decreasing at a particular point.

The connection is local. At duration $t$, a life aged $x$ at the starting point has attained age $x+t$. Consequently, the force that governs the slope of its survival function at that duration is $\mu_{x+t}$, not necessarily the force at the starting age.

Definition: Let $T_x$ be the future lifetime of a life aged $x$, and let $S_x(t)=P(T_x>t)$ be its survival function. Where the limit exists and $S_x(t)>0$, the force of mortality at duration $t$ is $$ \mu_{x+t} = \lim_{h\to 0^+} \frac{P(t<T_x\le t+h\mid T_x>t)}{h}. $$ It is an instantaneous rate, measured in inverse units of time. For example, when $t$ is measured in years, $\mu_{x+t}$ is measured per year.

The numerator is a conditional probability: among lives that have survived to duration $t$, it measures the probability of dying during the next $h$ units of time. The force is the limiting conditional probability per unit time as $h$ becomes small. This conditional perspective is essential. The unconditional probability of death during that interval also depends on the probability of having survived to its beginning.

Deriving the Survival Differential Equation

Assume a continuous lifetime model in which $S_x(t)$ is differentiable at the duration under consideration and positive there. Conditional on survival to $t$, the event of dying by $t+h$ has probability $P(t<T_x\le t+h\mid T_x>t)$. By the definition of conditional probability, this is the probability of surviving to $t$ and not surviving beyond $t+h$, divided by the probability of surviving to $t$.

P(t<T_x\le t+h\mid T_x>t) = \frac{S_x(t)-S_x(t+h)}{S_x(t)}.

Divide by $h$ and take the limit from the right. Differentiability gives the limiting difference quotient for $S_x$. The negative sign appears because survival decreases as duration increases.

\begin{aligned} \mu_{x+t} &= \lim_{h\to 0^+} \frac{S_x(t)-S_x(t+h)}{hS_x(t)}\\ &= -\frac{S_x'(t)}{S_x(t)}. \end{aligned}

Multiplying by $S_x(t)$ gives the differential equation. It states that the absolute rate of decrease of survival is the current survival probability multiplied by the current force.

Theorem: For a differentiable survival function with $S_x(t)>0$, the force of mortality and the slope of survival satisfy $$ \frac{d}{dt}S_x(t)=S_x'(t)=-\mu_{x+t}S_x(t). $$ The force is nonnegative, so the slope is nonpositive: survival cannot increase as duration increases.

The equation also distinguishes an absolute slope from a relative rate of decrease. The absolute slope $S_x'(t)$ depends on both the force and the amount of survival remaining. Dividing by $S_x(t)$ gives the relative slope $S_x'(t)/S_x(t)=-\mu_{x+t}$. Thus two survival curves can have different absolute slopes even when their forces are equal, if their survival probabilities differ.

Specify the duration.
At duration $t$, the life has attained age $x+t$.

Use the force at that attained age.
Evaluate $\mu_{x+t}$ using the model and the time units specified.

Multiply by survival.
Compute $S_x'(t)=-\mu_{x+t}S_x(t)$, retaining the negative sign.

Using the Equation at a Specified Point

The equation is most directly applied when a problem supplies the survival probability and the force at a particular duration. A force value alone does not give the absolute slope: the survival probability at that point is also needed. Conversely, the resulting slope is a local quantity; it describes the curve at the stated duration, not over an entire interval.

Worked Example: Finding the Slope at an Attained Age

In a hypothetical continuous lifetime model, a life is age 50 at the starting point. At duration $t=4$ years, the survival probability is $S_{50}(4)=0.80$, and the force at attained age 54 is $0.025$ per year. Find the slope of the survival function at that duration. State the method and identify a common mistake it avoids.

Method. Apply $S_x'(t)=-\mu_{x+t}S_x(t)$. The starting age is $x=50$, so at $t=4$ the attained age is $50+4=54$. The supplied force is therefore the appropriate force.

Substitute the values. $$ S_{50}'(4) = -\mu_{54}S_{50}(4) = -(0.025)(0.80) = -0.020\ \text{per year}. $$

The negative result means that survival is decreasing at this point. The force is a rate per year and the survival probability is dimensionless, so the slope is also expressed per year. This method avoids treating the force itself, $0.025$, as the slope; that would omit the factor $S_{50}(4)$.

Check. Dividing the calculated slope by survival gives $-0.020/0.80=-0.025$ per year, which is $-\mu_{54}$, as required.

Worked Example: Evaluating a Force That Varies With Age

A hypothetical continuous lifetime model specifies the force at attained age $y$ as $\mu_y=0.0005(y-30)$ per year. A life starts at age 50, and its survival probability at duration 4 years is $0.70$. Find the slope of its survival function at duration 4. Name the mistake that can result from using the wrong age.

Method. Evaluate the age-dependent force at the attained age first, then multiply by survival with a negative sign. At duration 4, attained age is $50+4=54$, so $$ \mu_{54}=0.0005(54-30)=0.0005(24)=0.012\ \text{per year}. $$

Now apply the differential equation: $$ S_{50}'(4)=-(0.012)(0.70)=-0.0084\ \text{per year}. $$

The survival curve is decreasing at $0.0084$ probability units per year at this point. The calculation uses the force at age 54, rather than substituting the starting age 50 into the force formula. Using age 50 would give $0.0005(20)=0.010$ per year and an incorrect slope of $-0.0070$ per year.

Check. The slope divided by the survival probability is $-0.0084/0.70=-0.012$ per year, matching the negative of the force evaluated at age 54.

Worked Example: Comparing Slopes for Two Hypothetical Lives

At a particular duration, two hypothetical continuous lifetime models give the following values: Model A has survival $0.60$ and force $0.030$ per year; Model B has survival $0.90$ and force $0.020$ per year. Calculate each local slope and determine which survival curve is decreasing more steeply in absolute value. State the common mistake avoided by comparing the calculated slopes.

Method. Use the same relationship for each model, pairing its own force with its own survival probability.

For Model A, $$ S_A' = -(0.030)(0.60)=-0.018\ \text{per year}. $$ For Model B, $$ S_B' = -(0.020)(0.90)=-0.018\ \text{per year}. $$

The slopes are equal, so neither curve is decreasing more steeply in absolute value at the specified point. Model A has the larger force, but its smaller survival probability offsets that difference in the absolute slope.

Check. The relative slopes are $-0.018/0.60=-0.030$ per year for Model A and $-0.018/0.90=-0.020$ per year for Model B. These equal the negatives of their respective forces. Comparing forces alone would incorrectly imply that Model A must have the more negative absolute slope.

Common Errors and Model Conditions

The differential equation is a local relationship, and correct use depends on identifying the point, the attained age, and the units. It applies where the survival function is differentiable and positive and the force is defined. If survival is zero, the conditional definition of force at that duration is not available because conditioning on survival has probability zero. If a survival curve has a nondifferentiable point, this derivative equation need not hold there in the ordinary pointwise sense.

Using the starting age instead of the attained age. At duration $t$, use $\mu_{x+t}$. The force need not be the same at ages $x$ and $x+t$.
Omitting survival. The slope is $-\mu_{x+t}S_x(t)$, not just $-\mu_{x+t}$.
Dropping the negative sign. A positive force corresponds to a nonpositive survival slope.
Mixing time units. If duration is measured in months, a force stated per year must be converted to a consistent time unit before calculating a slope per month.
Confusing local slope with an interval probability. The equation gives the derivative at one duration. It does not, by itself, give the exact probability of death over a nonzero interval.

Key takeaway: At duration $t$, survival is decreasing at the local rate $\mu_{x+t}S_x(t)$. Use the force at attained age $x+t$, multiply by survival at that duration, and attach a negative sign to obtain the slope.

Check Your Understanding

For each problem, use the survival differential equation at the stated duration. Keep age and time units consistent.

A hypothetical continuous lifetime model has $S_{35}(6)=0.92$ and force $0.015$ per year at attained age 41. Find the local slope of $S_{35}$. (Answer: $-0.0138$ per year)
A model specifies $\mu_y=0.0005(y-30)$ per year. For a life starting at age 50, the survival probability at duration 4 is $0.70$. Find the local slope at that duration. (Answer: $-0.0084$ per year)
At a specified point, a hypothetical model has survival $0.64$ and force $0.030$ per year. Find the slope and state whether survival is increasing or decreasing. (Answer: $-0.0192$ per year; decreasing)
Two models have, at their respective specified points, $(S,\mu)=(0.50,0.024)$ and $(0.80,0.015)$, with forces measured per year. Calculate both slopes and compare their absolute values. (Answer: both slopes are $-0.012$ per year, so their absolute values are equal)
A candidate calculates a slope of $-0.018$ per year when survival is $0.80$. The force at the attained age is $0.025$ per year. Is the candidate’s slope consistent with the model? Give the slope required by the differential equation. (Answer: No; the required slope is $-(0.025)(0.80)=-0.020$ per year)

What You'll Learn